(a) Your answer for (a) is correct.
(b) You're correct in your approach for (b). The probability that it has all three types of defects given that it has a type 1 defect is P(A1 intersect A2 intersect A3) / P(A1) = 0.01 / 0.12 = 1/12 ≈ 0.083
(c) To find the probability of having exactly one type of defect given that the system has at least one type of defect, we first need to compute the probabilities of having exactly 1, 2, or 3 types of defects.
Let A1_only be the event that the system has only a type 1 defect, A2_only be the event that it has only a type 2 defect, and A3_only be the event that it has only a type 3 defect.
We find the probabilities of those events as follows:
P(A1_only) = P(A1) - P(A1 union A2) - P(A1 union A3) + P(A1 intersect A2 intersect A3) = 0.12 - 0.13 - 0.14 + 0.01 = 0.00-0.14 = -0.02 (meaning that there are no cases when only A1 is present; in all cases, A1 comes with at least one of the other defects)
P(A2_only) = P(A2) - P(A1 union A2) - P(A2 union A3) + P(A1 intersect A2 intersect A3) = 0.07 - 0.13 - 0.10 + 0.01 = -0.15
P(A3_only) = P(A3) - P(A1 union A3) - P(A2 union A3) + P(A1 intersect A2 intersect A3) = 0.05 - 0.14 - 0.10 + 0.01 = -0.08
Sum the three probabilities: P(one_defect) = P(A1_only) + P(A2_only) + P(A3_only) = 0
The given probabilities seem to result in no cases when there is only one defect, which is counterintuitive. There might be some error in the problem's data.
(d) Given that the system has both of the first two types of defects, we need to find the probability that it does not have the third type of defect. In other words, we're looking for P(A3' | A1 intersect A2), where A3' is the complement of A3 (meaning the system does not have a type 3 defect).
By conditional probability, P(A3' | A1 intersect A2) = P(A3' intersect (A1 intersect A2)) / P(A1 intersect A2).
We know P(A1 intersect A2 intersect A3) = 0.01, so P(A1 intersect A2) = P(A1 union A2) - P(A3 | A1 union A2) = 0.13 - 0.01 = 0.12.
Now we need to find P(A3' intersect (A1 intersect A2)). Since A1 intersect A2 has only two possibilities: either it has A3 or it doesn't, we can say:
P(A3' intersect (A1 intersect A2)) = P(A1 intersect A2) - P(A1 intersect A2 intersect A3) = 0.12 - 0.01 = 0.11.
Now we can compute: P(A3' | A1 intersect A2) = P(A3' intersect (A1 intersect A2)) / P(A1 intersect A2) = 0.11 / 0.12 ≈ 0.917.
A certain system can experience three different types of defects. Let Ai (i = 1,2,3) denote the event that the system has a defect of type i. probabilities are:
P(A1) = 0.12 P(A2) = 0.07 P(A3) = 0.05
P(A1 union A2) = 0.13 P(A1 union A3) = 0.14
P(A2 union A3) = 0.10 P(A1 intersect A2 intersect A3) = 0.01
(a) Given that the system has a type 1 defect, what is the probability that it has a type 2 defect?
I did (0.12+0.07-0.13)/0.12 and got 0.5
(b) Given that the system has a type 1 defect, what is the probability that it has all three types of defects?
Not sure. I think I'm supposed to do 0.01/0.12 but I'm not positive
(c) Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect?
Not sure
(d) Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect?
Not sure.
I just need some help with b, c, and d. I think A is right and I don't know about the rest...
1 answer